Absolute value of sum of absolute values Wondering if it's generally true that:
$$||a|+| b||=|a|+|b|$$
If I assumme the contrary, that is $||a|+| b|| = m$, $|a|+|b|=k$, and $k \neq m$,
then $|a|+| b| = m$ or $|a|+| b| = -m$. But since $k \neq m$, this is a contradiction. So it has to be the case that they're equal.
Does this reasoning make sense ?
 A: For any real number $x,$ the absolute value or modulus of $x$ is defined as
$$|x| = \left\{
   \begin{array}{rl}
     x, & \text{if }  x \geq 0 \\
     -x, & \text{if } x < 0
   \end{array}\right.$$
Letting $x=|a|+|b|$ we see that since $|a|,|b|\ge 0$
$$|x| = ||a|+|b||=x=|a|+|b|$$
because both $|a|\ge 0$ and $|b|\ge 0$.
A: In the most naive sense, you could think of an absolute value as being the first square root of the square of a number. For example, $$|a| = \sqrt{a^2}$$
Then you could substitute and solve as follows:
$$||a|+|b||=|a|+|b|$$
$$\sqrt{(|a|+|b|)^2}=|a|+|b|$$
$$\sqrt{(\sqrt{a^2}+\sqrt{b^2})^2}=\sqrt{a^2}+\sqrt{b^2}$$
$$(\sqrt{a^2}+\sqrt{b^2})^2=(\sqrt{a^2}+\sqrt{b^2})^2$$
$$\sqrt{a^2}+\sqrt{b^2}=\sqrt{a^2}+\sqrt{b^2}$$
So I would say this generally holds true. I can't think of circumstance when it wouldn't -- even in the case of imaginary numbers, we seem to make out OK:
$$|a| = ||a||$$
$$\sqrt{a^2} = \sqrt{\sqrt{a^2}^2}$$
$$\sqrt{(1 + i)^2} = \sqrt{\sqrt{(1 + i)^2}^2}$$
$$\sqrt{1 + 2i - 1} = \sqrt{\sqrt{1 + 2i - 1}^2}$$
$$\sqrt{2i} = \sqrt{\sqrt{2i}^2}$$
$$\sqrt{2i} = \sqrt{(\sqrt{2}\sqrt{i})^2}$$
$$\sqrt{2i} = \sqrt{(\sqrt{2}\sqrt{i})(\sqrt{2}\sqrt{i})}$$
$$\sqrt{2i} = \sqrt{2i}$$
